Question

find the sum of n terms of the series : n+ 1 /n , n + 3/n , n + 5/n.....

Answer 1

Question;

find the sum of n terms of the series;

$\sf \: n + \frac{1}{n} , \: n + \frac{3}{n} ,n + \frac{5}{n} ..........n$


Sum of first terms;


$\sf \: n,n,n.....n + \frac{1}{n} (1 + 3 + 5) \\ \\ \\ \\ \tt \: sn = \frac{n}{2}(2a + (n - 1)d \\ \\ \sf \: sn = \frac{n}{2} (2n + (n - 1)0 \\ \\ \\ \sf \: sn = \frac{n}{2} (2n) \\ \\ \sf \: sn = \frac{ {2n}^{2} }{2} \\ \\ \sf \: \: sn = {n}^{2}$


Sum of first terms;



$\tt \: sn = \frac{n}{2}(2a + (n - 1) \\ \\ \sf \: sn = \frac{n}{2}(2 \times 1 + (n - 1)2 \\ \\ \sf \: sn = \frac{n}{2} (2 + 2n - 2) \\ \\ \\ \sf \: sn = \frac{n}{2} ( 2n )\\ \\ \\ \sf \: sn = \frac{ {2n}^{2} }{2} \\ \\ \sf \: sn = {n}^{2}$



According to the Question;



$\sf \: {n}^{2} + \frac{1}{n} ( {n}^{2} ) \\ \\ \tt \: {n}^{2} + \frac{ {n}^{2} }{n} \\ \\ \\ \sf \: \frac{ {n}^{3} + {n}^{2} }{n} \\ \ \\ \\ \sf \: \frac{ {n}^{2}(n + 1) }{n} \\ \\ \sf \: \frac{ { \cancel{n}}^{2}(n + 1) }{ \cancel{n}} \\ \\ \\ \sf \: n(n + 1)$



Hence, find the sum of n terms of the series;



$\sf \: n + \frac{1}{n} , \: n + \frac{3}{n} ,n + \frac{5}{n} ..........n = n(n + 1)$

Answer 2

ANSWER :-

n² + n


EXPLAINATION :-

IN THE ATTACHMENT